# series involving $(-1)^k$

Given the series ${a_n}$ : $a_n= \sum _{k=1}^{n}{\frac { \left( -1 \right) ^{k}}{{3}^{k}+\sqrt {k}}}$ prove that ${a_n}$ converges.

So far this is what I've done: I've split this into 2 sub-series: $k=2n$ and $k=2n+1$

for $(k=2n)$ I get: $\sum _{k=1}^{n} \left( {3}^{k}+\sqrt {k} \right) ^{-1} = 0$ (as k goes to infinity the whole series converges to 0)

for $(k=2n+1)$ I get:

$\sum _{k=1}^{n}- \left( \left( {3}^{k} \right) ^{-1}+\sqrt {-k} \right) ^{-1}$ (which also converges to 0 as k approaches to infinity)

Since both subs-series of $a_n$ converge to the same value, $0$, $a_n$ converges, in particular to $0$.

I'm not sure if this is a tight enough proof since maybe by chance it works out here specifically. I also checked separate limits as k approaches $-\infty$. They both converge to $0$ for each sub-series. What am I missing? Thanks.

-
Don't write $k=2n$ when the letter $n$ is already in use. Additionally, don't write $\sum_{k=1}^n$ if you mean to only sum over odds or evens in particular. Lastly: (a) how can a sum of positive numbers equal $0$, and (b) how are you getting negative $k$ in your sum over odd $k$? – anon Feb 2 '12 at 7:55

$$|a_n| \le \sum_{k=1}^n \frac{ 1}{3^k + \sqrt k} \le \sum_{k=1}^n \frac{1}{3^k - \frac 12 3^k} < 2 \sum_{k=1}^n \left( \frac 13 \right)^k$$ which converges because this is a geometric series. Since $a_n$ converges absolutely it converges conditionally.

You could also use Leibniz's criterion, which states that if a series has leading term $(-1)^k b_k$ with $b_k \to 0$, then $\sum (-1)^k b_k$ converges. The fact that $b_k = \frac 1{3^k + \sqrt k} \to 0$ is clear enough.

EDIT : As the comments pointed out, I didn't remember at the time all the hypothesis of Leibniz's Criterion, which are that on top of what I said, the $b_k$'s must be, up to some $N$, non-negative and decreasing.

Hope that helps,

-
thanks, Leibniz's criterion makes this very clear. Glad to discover it. – nofe Feb 2 '12 at 8:08
The alternating series test (aka Leibniz criterion) applies when the terms have alternating sign, converge to zero and are monotone decreasing (in absolute value). This last part is essential and you can construct counterexamples if you leave it out. – Najib Idrissi Feb 2 '12 at 13:22
It would also be good to emphasize that the $b_k$ are nonnegative (this is implied by the hypotheses that @zulon adds), since you could do something silly like $b_k = (-1)^k(1/k)$. I'm interested in seeing an example in which $0 \leq b_k \to 0$ and yet $b_k$ not being monotone messes things up. – Dylan Moreland Feb 2 '12 at 16:01
Take $u_n = {1 \over n}$ for even $n$ and $u_n = {-1 \over 2n}$ for odd $n$. Then $u_n$ converges to zero, is alternating (but not monotone in absolute value), and $\sum u_n$ diverges. Even simpler take $u_n = {(-1)^n \over \sqrt{n} + (-1)^n}$, then $\sum u_n$ diverges (easily seen using an asymptotic expansion). – Najib Idrissi Feb 2 '12 at 18:00
Okay guys, sorry if I didn't remember all the hypothesis. It's good that you pointed them out, but my point was mostly that you could use the criterion, not that I wanted to describe it completely. – Patrick Da Silva Feb 2 '12 at 19:56

You can use Leibniz's Rule for alternating series.

If $\{ a_n\}$ is a monotonic decrasing sucesion of numbers, the alternate series converges, and if $S$ is the sum, then, for every $n > 0$: $$0 < (-1)^n (S-s_n) < a_{n+1}$$ and $$0 < S < a_1$$

Thus your series converges, and you can calculate it's sum approximately with the first inequality.

-